3.112 \(\int \frac {x^3 (a+b \tanh ^{-1}(c x))^2}{(d+c d x)^3} \, dx\)

Optimal. Leaf size=337 \[ -\frac {3 b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}-\frac {11 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (c x+1)}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (c x+1)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (c x+1)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (c x+1)^2}+\frac {19 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^4 d^3}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}+\frac {3 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^4 d^3}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c^4 d^3}-\frac {21 b^2}{16 c^4 d^3 (c x+1)}+\frac {b^2}{16 c^4 d^3 (c x+1)^2}+\frac {21 b^2 \tanh ^{-1}(c x)}{16 c^4 d^3} \]

[Out]

1/16*b^2/c^4/d^3/(c*x+1)^2-21/16*b^2/c^4/d^3/(c*x+1)+21/16*b^2*arctanh(c*x)/c^4/d^3+1/4*b*(a+b*arctanh(c*x))/c
^4/d^3/(c*x+1)^2-11/4*b*(a+b*arctanh(c*x))/c^4/d^3/(c*x+1)+19/8*(a+b*arctanh(c*x))^2/c^4/d^3+x*(a+b*arctanh(c*
x))^2/c^3/d^3+1/2*(a+b*arctanh(c*x))^2/c^4/d^3/(c*x+1)^2-3*(a+b*arctanh(c*x))^2/c^4/d^3/(c*x+1)-2*b*(a+b*arcta
nh(c*x))*ln(2/(-c*x+1))/c^4/d^3+3*(a+b*arctanh(c*x))^2*ln(2/(c*x+1))/c^4/d^3-b^2*polylog(2,1-2/(-c*x+1))/c^4/d
^3-3*b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/c^4/d^3-3/2*b^2*polylog(3,1-2/(c*x+1))/c^4/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.66, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 31, number of rules used = 14, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {5940, 5910, 5984, 5918, 2402, 2315, 5928, 5926, 627, 44, 207, 5948, 6056, 6610} \[ -\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}-\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^4 d^3}-\frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 c^4 d^3}-\frac {11 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (c x+1)}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (c x+1)^2}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (c x+1)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (c x+1)^2}+\frac {19 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^4 d^3}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}+\frac {3 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}-\frac {21 b^2}{16 c^4 d^3 (c x+1)}+\frac {b^2}{16 c^4 d^3 (c x+1)^2}+\frac {21 b^2 \tanh ^{-1}(c x)}{16 c^4 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcTanh[c*x])^2)/(d + c*d*x)^3,x]

[Out]

b^2/(16*c^4*d^3*(1 + c*x)^2) - (21*b^2)/(16*c^4*d^3*(1 + c*x)) + (21*b^2*ArcTanh[c*x])/(16*c^4*d^3) + (b*(a +
b*ArcTanh[c*x]))/(4*c^4*d^3*(1 + c*x)^2) - (11*b*(a + b*ArcTanh[c*x]))/(4*c^4*d^3*(1 + c*x)) + (19*(a + b*ArcT
anh[c*x])^2)/(8*c^4*d^3) + (x*(a + b*ArcTanh[c*x])^2)/(c^3*d^3) + (a + b*ArcTanh[c*x])^2/(2*c^4*d^3*(1 + c*x)^
2) - (3*(a + b*ArcTanh[c*x])^2)/(c^4*d^3*(1 + c*x)) - (2*b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(c^4*d^3) +
(3*(a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/(c^4*d^3) - (b^2*PolyLog[2, 1 - 2/(1 - c*x)])/(c^4*d^3) - (3*b*(a
+ b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(c^4*d^3) - (3*b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*c^4*d^3)

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 5926

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b
*ArcTanh[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 5928

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(
a + b*ArcTanh[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^(p
- 1), (d + e*x)^(q + 1)/(1 - c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] &
& NeQ[q, -1]

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] - Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[2, 1 - u])
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2
/(1 + c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{(d+c d x)^3} \, dx &=\int \left (\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)^3}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}\right ) \, dx\\ &=\frac {\int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c^3 d^3}-\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^3} \, dx}{c^3 d^3}+\frac {3 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx}{c^3 d^3}-\frac {3 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1+c x} \, dx}{c^3 d^3}\\ &=\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {b \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^3}+\frac {a+b \tanh ^{-1}(c x)}{4 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}+\frac {(6 b) \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}-\frac {(6 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^3 d^3}-\frac {(2 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c^2 d^3}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 c^3 d^3}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 c^3 d^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 c^3 d^3}-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c^3 d^3}+\frac {(3 b) \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^3 d^3}-\frac {(3 b) \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{c^3 d^3}+\frac {\left (3 b^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^3 d^3}\\ &=\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)^2}-\frac {11 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)}+\frac {19 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^4 d^3}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^4 d^3}-\frac {b^2 \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{4 c^3 d^3}-\frac {b^2 \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 c^3 d^3}+\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^3 d^3}+\frac {\left (3 b^2\right ) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^3 d^3}\\ &=\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)^2}-\frac {11 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)}+\frac {19 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^4 d^3}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^4 d^3}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c^4 d^3}-\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{4 c^3 d^3}-\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{4 c^3 d^3}+\frac {\left (3 b^2\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{c^3 d^3}\\ &=\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)^2}-\frac {11 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)}+\frac {19 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^4 d^3}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^4 d^3}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^4 d^3}-\frac {b^2 \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c^3 d^3}-\frac {b^2 \int \left (\frac {1}{2 (1+c x)^3}+\frac {1}{4 (1+c x)^2}-\frac {1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c^3 d^3}+\frac {\left (3 b^2\right ) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}\\ &=\frac {b^2}{16 c^4 d^3 (1+c x)^2}-\frac {21 b^2}{16 c^4 d^3 (1+c x)}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)^2}-\frac {11 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)}+\frac {19 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^4 d^3}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^4 d^3}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^4 d^3}+\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{16 c^3 d^3}+\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{8 c^3 d^3}-\frac {\left (3 b^2\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{2 c^3 d^3}\\ &=\frac {b^2}{16 c^4 d^3 (1+c x)^2}-\frac {21 b^2}{16 c^4 d^3 (1+c x)}+\frac {21 b^2 \tanh ^{-1}(c x)}{16 c^4 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)^2}-\frac {11 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^4 d^3 (1+c x)}+\frac {19 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^4 d^3 (1+c x)^2}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^4 d^3}+\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^4 d^3}-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^4 d^3}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^4 d^3}\\ \end {align*}

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Mathematica [A]  time = 1.30, size = 418, normalized size = 1.24 \[ \frac {64 a^2 c x-\frac {192 a^2}{c x+1}+\frac {32 a^2}{(c x+1)^2}-192 a^2 \log (c x+1)+4 a b \left (16 \log \left (1-c^2 x^2\right )-48 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+20 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )-20 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (8 c x+24 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+10 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )-10 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )\right )\right )+b^2 \left (-64 \left (3 \tanh ^{-1}(c x)-1\right ) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )-96 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )+64 c x \tanh ^{-1}(c x)^2-64 \tanh ^{-1}(c x)^2+192 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-128 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+80 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )-8 \tanh ^{-1}(c x)^2 \sinh \left (4 \tanh ^{-1}(c x)\right )+80 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )-4 \tanh ^{-1}(c x) \sinh \left (4 \tanh ^{-1}(c x)\right )+40 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )-80 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )+8 \tanh ^{-1}(c x)^2 \cosh \left (4 \tanh ^{-1}(c x)\right )-80 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \cosh \left (4 \tanh ^{-1}(c x)\right )-40 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )\right )}{64 c^4 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*(a + b*ArcTanh[c*x])^2)/(d + c*d*x)^3,x]

[Out]

(64*a^2*c*x + (32*a^2)/(1 + c*x)^2 - (192*a^2)/(1 + c*x) - 192*a^2*Log[1 + c*x] + 4*a*b*(-20*Cosh[2*ArcTanh[c*
x]] + Cosh[4*ArcTanh[c*x]] + 16*Log[1 - c^2*x^2] - 48*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 20*Sinh[2*ArcTanh[c*x
]] + 4*ArcTanh[c*x]*(8*c*x - 10*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 24*Log[1 + E^(-2*ArcTanh[c*x])]
+ 10*Sinh[2*ArcTanh[c*x]] - Sinh[4*ArcTanh[c*x]]) - Sinh[4*ArcTanh[c*x]]) + b^2*(-64*ArcTanh[c*x]^2 + 64*c*x*A
rcTanh[c*x]^2 - 40*Cosh[2*ArcTanh[c*x]] - 80*ArcTanh[c*x]*Cosh[2*ArcTanh[c*x]] - 80*ArcTanh[c*x]^2*Cosh[2*ArcT
anh[c*x]] + Cosh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*Cosh[4*ArcTanh[c*x]] + 8*ArcTanh[c*x]^2*Cosh[4*ArcTanh[c*x]]
- 128*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + 192*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] - 64*(-1 +
3*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] - 96*PolyLog[3, -E^(-2*ArcTanh[c*x])] + 40*Sinh[2*ArcTanh[c*x
]] + 80*ArcTanh[c*x]*Sinh[2*ArcTanh[c*x]] + 80*ArcTanh[c*x]^2*Sinh[2*ArcTanh[c*x]] - Sinh[4*ArcTanh[c*x]] - 4*
ArcTanh[c*x]*Sinh[4*ArcTanh[c*x]] - 8*ArcTanh[c*x]^2*Sinh[4*ArcTanh[c*x]]))/(64*c^4*d^3)

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{3} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x^{3} \operatorname {artanh}\left (c x\right ) + a^{2} x^{3}}{c^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

integral((b^2*x^3*arctanh(c*x)^2 + 2*a*b*x^3*arctanh(c*x) + a^2*x^3)/(c^3*d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d^3*x
+ d^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{3}}{{\left (c d x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2*x^3/(c*d*x + d)^3, x)

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maple [C]  time = 0.96, size = 5750, normalized size = 17.06 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a^{2} {\left (\frac {6 \, c x + 5}{c^{6} d^{3} x^{2} + 2 \, c^{5} d^{3} x + c^{4} d^{3}} - \frac {2 \, x}{c^{3} d^{3}} + \frac {6 \, \log \left (c x + 1\right )}{c^{4} d^{3}}\right )} + \frac {{\left (2 \, b^{2} c^{3} x^{3} + 4 \, b^{2} c^{2} x^{2} - 4 \, b^{2} c x - 5 \, b^{2} - 6 \, {\left (b^{2} c^{2} x^{2} + 2 \, b^{2} c x + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{8 \, {\left (c^{6} d^{3} x^{2} + 2 \, c^{5} d^{3} x + c^{4} d^{3}\right )}} - \int -\frac {{\left (b^{2} c^{4} x^{4} - b^{2} c^{3} x^{3}\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c^{4} x^{4} - a b c^{3} x^{3}\right )} \log \left (c x + 1\right ) - {\left (2 \, {\left (2 \, a b c^{4} + b^{2} c^{4}\right )} x^{4} - 9 \, b^{2} c x - 2 \, {\left (2 \, a b c^{3} - 3 \, b^{2} c^{3}\right )} x^{3} - 5 \, b^{2} + 2 \, {\left (b^{2} c^{4} x^{4} - 4 \, b^{2} c^{3} x^{3} - 9 \, b^{2} c^{2} x^{2} - 9 \, b^{2} c x - 3 \, b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \, {\left (c^{7} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{3} - 2 \, c^{4} d^{3} x - c^{3} d^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

-1/2*a^2*((6*c*x + 5)/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - 2*x/(c^3*d^3) + 6*log(c*x + 1)/(c^4*d^3)) + 1/8*
(2*b^2*c^3*x^3 + 4*b^2*c^2*x^2 - 4*b^2*c*x - 5*b^2 - 6*(b^2*c^2*x^2 + 2*b^2*c*x + b^2)*log(c*x + 1))*log(-c*x
+ 1)^2/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - integrate(-1/4*((b^2*c^4*x^4 - b^2*c^3*x^3)*log(c*x + 1)^2 + 4*
(a*b*c^4*x^4 - a*b*c^3*x^3)*log(c*x + 1) - (2*(2*a*b*c^4 + b^2*c^4)*x^4 - 9*b^2*c*x - 2*(2*a*b*c^3 - 3*b^2*c^3
)*x^3 - 5*b^2 + 2*(b^2*c^4*x^4 - 4*b^2*c^3*x^3 - 9*b^2*c^2*x^2 - 9*b^2*c*x - 3*b^2)*log(c*x + 1))*log(-c*x + 1
))/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*atanh(c*x))^2)/(d + c*d*x)^3,x)

[Out]

int((x^3*(a + b*atanh(c*x))^2)/(d + c*d*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x^{3}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b^{2} x^{3} \operatorname {atanh}^{2}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {2 a b x^{3} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c*x))**2/(c*d*x+d)**3,x)

[Out]

(Integral(a**2*x**3/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b**2*x**3*atanh(c*x)**2/(c**3*x**3 +
3*c**2*x**2 + 3*c*x + 1), x) + Integral(2*a*b*x**3*atanh(c*x)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x))/d**3

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